Problem: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $z \neq 0$. $r = \dfrac{z + 3}{5z^2 - 25z - 120} \times \dfrac{-5z^2 - 55z - 150}{z + 6} $
Explanation: First factor out any common factors. $r = \dfrac{z + 3}{5(z^2 - 5z - 24)} \times \dfrac{-5(z^2 + 11z + 30)}{z + 6} $ Then factor the quadratic expressions. $r = \dfrac {z + 3} {5(z + 3)(z - 8)} \times \dfrac {-5(z + 6)(z + 5)} {z + 6} $ Then multiply the two numerators and multiply the two denominators. $r = \dfrac {(z + 3) \times -5(z + 6)(z + 5) } { 5(z + 3)(z - 8) \times (z + 6)} $ $r = \dfrac {-5(z + 6)(z + 5)(z + 3)} {5(z + 3)(z - 8)(z + 6)} $ Notice that $(z + 3)$ and $(z + 6)$ appear in both the numerator and denominator so we can cancel them. $r = \dfrac {-5(z + 6)(z + 5)\cancel{(z + 3)}} {5\cancel{(z + 3)}(z - 8)(z + 6)} $ We are dividing by $z + 3$ , so $z + 3 \neq 0$ Therefore, $z \neq -3$ $r = \dfrac {-5\cancel{(z + 6)}(z + 5)\cancel{(z + 3)}} {5\cancel{(z + 3)}(z - 8)\cancel{(z + 6)}} $ We are dividing by $z + 6$ , so $z + 6 \neq 0$ Therefore, $z \neq -6$ $r = \dfrac {-5(z + 5)} {5(z - 8)} $ $ r = \dfrac{-(z + 5)}{z - 8}; z \neq -3; z \neq -6 $